Problem: We have a triangle $\triangle ABC$ and a point $K$ on $BC$ such that $AK$ is an altitude of $\triangle ABC$. If $AC = 10,$ $BK = 7$, and $BC = 13,$ then what is the area of $\triangle ABC$?
Solution: First, we sketch! [asy]
pair A, B, C, K;
A = (0, 8);
B = (-7, 0);
C = (6, 0);
K = (0, 0);
draw(A--B--C--cycle);
draw(A--K);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$K$", K, NE);
label("10", C--A, NE);
label("7", B--K, N);
label("13", B--C, S);
draw(rightanglemark(A,K,B,10));
[/asy] We now see that $CK = BC - BK = 6.$ That means $\triangle AKC$ is a $3:4:5$ right triangle, so $AK = 8.$ At this point, we can see that the area of $\triangle ABC$ is $\frac{1}{2} \cdot AK \cdot BC = \frac{1}{2} \cdot 8 \cdot 13 = \boxed{52}.$